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The charge per unit length of four quadrant of the ring of radius R is 2 \lambda, -2 \lambda, 1...

Question:

The charge per unit length of four quadrant of the ring of radius {eq}R {/eq} is {eq}2 \lambda, -2 \lambda, 1 \lambda {/eq} and {eq}-1 \lambda {/eq} respectively. Find out the electric field at the center of the ring?

{eq}(a) -\frac{\lambda}{2 \pi \epsilon_R} \hat{i} \\ (b) \frac{\lambda}{2 \pi \epsilon_R} \hat{j} \\ (c) \frac{\sqrt{2} \lambda}{4 \pi \epsilon_R} \hat{j} \\ {/eq}

(d) None of the above

Electric Field:

In physics, the electric field is known as the region of space that takes place near a particle within which a force is exercised on other charged objects. Its value is expressed in the units of volts per meters.

Answer and Explanation:


Given Data

  • The charge per unit length of fist quadrant is: {eq}{\lambda _1} = 2\lambda {/eq}
  • The charge per unit length of second quadrant is: {eq}{\lambda _2} = - 2\lambda {/eq}
  • The charge per unit length of third quadrant is: {eq}{\lambda _3} = \lambda {/eq}
  • The charge per unit length of fourth quadrant is: {eq}{\lambda _1} = - \lambda {/eq}


Take a small element having charge dq for a quarter of ring and calculate the electric field for this element.

{eq}dE = \dfrac{{dq}}{{4\pi {\varepsilon _0}{R^2}}} {/eq}

Here, {eq}{\varepsilon _0} {/eq} is the permittivity of free space and R is the radius of the ring.


The component of electric field at the center along x-axis is as follows,

{eq}\begin{align*} {E_x} &= \int {dE\cos \theta } \\ {E_x} &= \int {\dfrac{{dq}}{{4\pi {\varepsilon _0}{R^2}}}\cos \theta } \\ {E_x} &= \int {\dfrac{{\lambda Rd\theta }}{{4\pi {\varepsilon _0}{R^2}}}\cos \theta } \\ {E_x} &= \int {\dfrac{\lambda }{{4\pi {\varepsilon _0}R}}\cos \theta } .d\theta \end{align*} {/eq}

Here, {eq}\theta {/eq} is angle between the quadrant and the horizontal axis.


Integrate over all four quadrants of the ring.

{eq}\begin{align*} {E_x} &= \dfrac{1}{{4\pi {\varepsilon _0}R}}\left( {{\lambda _1}\int\limits_0^{{\pi {\left/ {\vphantom {\pi 2}} \right. } 2}} {\cos \theta .d\theta } - {\lambda _2}\int\limits_{{\pi {\left/ {\vphantom {\pi 2}} \right. } 2}}^\pi {\cos \theta .d\theta } + {\lambda _3}\int\limits_\pi ^{{{3\pi } {\left/ {\vphantom {{3\pi } 2}} \right. } 2}} {\cos \theta .d\theta } - {\lambda _4}\int\limits_{{{3\pi } {\left/ {\vphantom {{3\pi } 2}} \right. } 2}}^{2\pi } {\cos \theta .d\theta } + } \right)\\ {E_x} &= \dfrac{1}{{4\pi {\varepsilon _0}R}}\left( {2\lambda \int\limits_0^{{\pi {\left/ {\vphantom {\pi 2}} \right. } 2}} {\cos \theta .d\theta } - 2\lambda \int\limits_{{\pi {\left/ {\vphantom {\pi 2}} \right. } 2}}^\pi {\cos \theta .d\theta } + \lambda \int\limits_\pi ^{{{3\pi } {\left/ {\vphantom {{3\pi } 2}} \right. } 2}} {\cos \theta .d\theta } - \lambda \int\limits_{{{3\pi } {\left/ {\vphantom {{3\pi } 2}} \right. } 2}}^{2\pi } {\cos \theta .d\theta } + } \right)\\ {E_x} &= \dfrac{1}{{4\pi {\varepsilon _0}r}}\left( {2\lambda \left( {1 - 0} \right) - 2\lambda \left( {0 - 1} \right) + \lambda \left( { - 1 - 0} \right) - \lambda \left( {0 + 1} \right)} \right)\\ {E_x} &= \dfrac{1}{{4\pi {\varepsilon _0}R}}\left( {2\lambda + 2\lambda - \lambda - \lambda } \right)\\ {E_x} &= \dfrac{{2\lambda }}{{4\pi {\varepsilon _0}R}}\\ {E_x} &= \dfrac{\lambda }{{2\pi {\varepsilon _0}R}} \end{align*} {/eq}


The component of electric field at the center along y-axis is as follows,

{eq}\begin{align*} {E_y} &= \int {dE\sin \theta } \\ {E_y} &= \int {\dfrac{{dq}}{{4\pi {\varepsilon _0}{R^2}}}\sin \theta } \\ {E_y} &= \int {\dfrac{{\lambda rd\theta }}{{4\pi {\varepsilon _0}{R^2}}}\sin \theta } \\ {E_y} &= \int {\dfrac{\lambda }{{4\pi {\varepsilon _0}R}}\sin } \;\theta .d\theta \end{align*} {/eq}


Integrate over all four quadrants of the ring.

{eq}\begin{align*} {E_y} &= \dfrac{1}{{4\pi {\varepsilon _0}R}}\left( {{\lambda _1}\int\limits_0^{{\pi {\left/ {\vphantom {\pi 2}} \right. } 2}} {\sin \theta .d\theta } - {\lambda _2}\int\limits_{{\pi {\left/ {\vphantom {\pi 2}} \right. } 2}}^\pi {\sin \theta .d\theta } + {\lambda _3}\int\limits_\pi ^{{{3\pi } {\left/ {\vphantom {{3\pi } 2}} \right. } 2}} {\sin \theta .d\theta } - {\lambda _4}\int\limits_{{{3\pi } {\left/ {\vphantom {{3\pi } 2}} \right. } 2}}^{2\pi } {sin\theta .d\theta } + } \right)\\ {E_y} &= \dfrac{1}{{4\pi {\varepsilon _0}R}}\left( {2\lambda \int\limits_0^{{\pi {\left/ {\vphantom {\pi 2}} \right. } 2}} {\sin \theta .d\theta } - 2\lambda \int\limits_{{\pi {\left/ {\vphantom {\pi 2}} \right. } 2}}^\pi {\sin \theta .d\theta } + \lambda \int\limits_\pi ^{{{3\pi } {\left/ {\vphantom {{3\pi } 2}} \right. } 2}} {\sin \theta .d\theta } - \lambda \int\limits_{{{3\pi } {\left/ {\vphantom {{3\pi } 2}} \right. } 2}}^{2\pi } {\sin \theta .d\theta } + } \right)\\ {E_y} &= \dfrac{1}{{4\pi {\varepsilon _0}R}}\left( 0 \right)\\ {E_y} &= 0 \end{align*} {/eq}


The net electric field at the center of the ring is as follows,

{eq}\begin{align*} E &= {E_x} + {E_y}\\ E &= \dfrac{\lambda }{{2\pi {\varepsilon _0}R}} + 0\\ E &= \dfrac{\lambda }{{2\pi {\varepsilon _0}R}} \end{align*} {/eq}


Thus, the correct answer is option (b).


Learn more about this topic:

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